Problems & Puzzles:
Conjectures
Conjecture 107. The
Shyam's conjecture about square-free triangular numbers
On Nov 29, 2024 Shyam Sunder Gupta sent
the following conjecture:
The nth triangular number, represented by Tn, equals
n*(n+1)/2.
Consider a set of three consecutive triangular numbers Tn,
T(n+1), and T(n+2), such that
the number of prime factors (counting multiplicities) of
each of the three consecutive
triangular numbers is three. The smallest set of such
three consecutive triangular
numbers is as follows:
T17 = 153 = 3 * 3 * 17, T18 = 171 = 3 * 3 * 19, and T19 =
190 = 2 * 5 * 19.
It can be seen that T19 is square-free, whereas T17 and
T18 are not square-free.
The next set of three consecutive triangular numbers,
each with three prime factors
(counting multiplicities), is (406, 435, 465) as
T28 = 406 = 2 * 7 * 29, T29 = 435 = 3 * 5 * 29, and T30 =
465 = 3 * 5 * 31.
It can be seen that T28, T29, and T30 are all
square-free.
Reference can be made to OEIS, where I submitted the
sequence (A349696) of
"Smallest number in a set of three consecutive triangular
numbers, each with three prime
factors (counted with multiplicity)." with a comment that
153 is the only known number in
the sequence that is not square-free.
Conjecture:
"For a set of three consecutive triangular numbers Tn,
T(n+1), and T(n+2), such that if the number of prime factors (counting
multiplicities) of each of the three consecutive triangular numbers is
three, then except for the smallest set, i.e., (153, 171, 190), all
three triangular numbers in all other sets are square-free."
(I
have tested it for triangular numbers < 10^17, and the conjecture
holds.)
1) Can this conjecture be proved or disproved?
2) In case this conjecture is disproved, can someone find
a counterexample?

From 21-27 Dec,. 2024 contributions came from Fred Schneider,
Alain Rochelli, Michael Branicky, Alessandro Casini, Emmanuel
Vantieghem
***
Fred wrote:
First consider the formulas of three consecutive triangular numbers.
T(n) = n * (n + 1) /2
T(n+1) = (n+ 1) * (n + 2) /2
T(n+1) = (n+ 2) * (n + 3) /2
These three triangular numbers contain the factors of n through
n + 3 (barring a single 2 in the denominator). Note that among
4 consecutive numbers, one has to be a multiple of 4, two have
to be even, and at least one has to be a multiple of 3.
The convention used below is that p, p1, p2 and p3 are all prime
numbers with p being the squared factor.
The non-square free solution had this form for the four
consecutive numbers:
n = p1
n+1 = 2 * p * p
n+2 = p2
n+3 = 4 * p3
resulting in the three consecutive triangular numbers, each with
exactly three prime factors:
T(n) = n * (n+1) / 2 = p1 * p * p
T(n+1) = (n+1) * (n+2) / 2 = p2 * p * p
T(n+2) = (n+2) * (n+3) / 2 = 2 * p2 * p3
where n = 17, p1 = 17, p2 = 19 , p = 3 and p3 = 5
A counterexample to this conjecture would either have to have the
above arrangement for the four consecutive numbers or one of these
other three arrangements.
(Arrangement #2)
n = p1
n+1 = 4 * p2
n+2 = p3
n+3 = 2 * p * p
(Arrangement #3)
n = 2 * p * p
n+1 = p1
n+2 = 4 * p2
n+ 3 = p3
(Arrangement #4)
n = 4 * p1
n+1 = p2
n+2 = 2 * p * p
n+ 3 = p3
As p is a squared prime factor, it
must be always less than p1, p2 and p3 in one of these
arrangements beyond a trivial threshold. (At n= 6, p = 2 and we
get non-prime p1 = 1.5. At n = 7, p = 2 and p2 = 2.5)
As stated before, each set of four
consecutive numbers must have one number divisible by 3. So,
either p is 3 or p = 2 and one of the other primes is 3.
If p = 2 then 2 * p * p = 8. This
is not possible because a multiple of 4 is always two away from
it in each arrangement.
So, p must be 3. As the known
solution is for the first arrangement, we check the remaining
arrangements:
2) n + 3 = 2 * p * p = 18, n = 15
(not a prime)
3) n = 2 * p * p = 18, n + 3 = 21
(not a prime)
4) n + 2 = 2 * p * p = 18, n = 16 which is of the form 4 * x.
but x = 4 is not a prime
So, there are no possible counterexamples and the conjecture is
true.
***
Alain wrote:
This conjecture is stated in A349696 and completed in A255200.
Furthermore, we have T(n)= A000217 ( A255200 (n)).
It is easy to prove that Tn is always divisible by 2 or 3.
The generic notation of T(n) is then T(n)=2*q*r or T(n)=3*p*r with
3<p<q<r if the conjecture is true. If it is false we have p=r or q=r.
First case: let T(n)=n*(n+1)/2=2*r^2
--> n*(n+1)=(2*r)^2 which is not possible with n integer and r prime
(i.e. n*(n+1) is never a square).
Second case: let T(n)=n*(n+1)/2=3*r^2
--> T(n+1)=(n+1)*(n+2)/2 = [n*(n+1)/2]*[(n+2)/n] = T(n)*(n+2)/n =
3*r^2*(n+2)/n
T(n+1) is 3-almost prime only if n=r with r and r+2 twin primes -->
T(n+1)=3*r*(r+2)
T(n+2)=(n+2)*(n+3)/2 = [(n+1)*(n+2)/2]*[(n+3)/(n+1)] =
T(n+1)*(n+3)/(n+1) = 3*r*(r+2)*(n+3)/(n+1)
With n=r --> T(n+2) = 3*r*(r+2)*(r+3)/(r+1) which is never a 3-almost
prime.
In conclusion, it seems that this conjecture is true so it is no longer
a conjecture.
***
Michael wrote:
I tested the conjecture up to T(867608681) = 376372412107084221 > 3.76 *
10^17.
A search from scratch that exceeds
that limit would take >24 hr on the same hardware.
***
Alessandro wrote:
The conjecture is true. Between (n, n+1, n+2, n+3), i.e. the numbers
that in pairs form the triangular numbers, one must be divisible by 4
(contributing a factor of 2 in triangular ones) e one (two in some
cases) is a multiple of 3. With these constraints, it's easy to see
that, except for a finite number of cases, the only ways of constructing
triangular numbers are (n, n+1, n+2, n+3) of the form (p, 6q, s, 4t) or
(4p, q, 6s, t), with p, q, s, t primes. Therefore, the only way to have
a no square-free triangular numbers is among the finite cases, in which
only n=17 appears.
***
Emmanuel wrote:
The conjecture is true !
We shall prove that there exist only
two consecutive triangular numbers that have three prime factors
and that are non square-free.
Proof.
Let's examine how a triangular number t can be a non square-free
product of three primes.
1) t = p^3
This is totally impossible since t has to be of the form
n(n+1)/2 :
Such numbers must have at least two different prime factors.
2) t = 4p, with p an odd prime.
Since 1 + 8t = 1 + 32p must be a square, say : x^2 we may
deduce :
32p = (x - 1)(x + 1).
The only common factor of x - 1 and x + 1 is 2.
So, one of them must be 2p and the other must be 16.
This can only be true if p = 7 or : t =28 = 7*8/2.
In that case, the former triangular is 21 (product of two
primes)
and the next triangular is 36 (four primes)
3) t = 2 p^2, with p an odd prime.
Then, 1 + 16 p^2 must be of the form x^2 and thus :
16p^2 = (x - 1)(x + 1).
The only possibility to achieve that is that one of the two
numbers x - 1, x + 1 must be 2 p^2
while the other must be 8,
totally impossible !
4) T = p q^2 with p, q odd primes.
Thus, we must have : 8 p q^2 = (x - 1)(x + 1). There are four
possibilities :
a) 4p = x + 1, 2q^2 = x - 1
or : 4q^2 = 4p - 2
or : 2q^2 = 2p - 1, completely impossible
b) 2p = x + 1, 4q^2 = x - 1
or : 4q^2 = 2p - 2
or : 2q^2 + 1 = p. Only p = 19, q = 3 seems to
satisfy this
(look at the equation modulo 3)
c) 4p = x - 1, 2q^2 = x + 1
or : 4q^2 = 4p + 2
or : 2q^2 = 2p + 1 ! Totally impossible
d) 2p = x - 1, 4q^2 = x + 1
or : 4q^2 = 2p + 2
or : 2q^2 - 1 = p. There are many many solutions : see https://oeis.org/A092057.
It is easy to show that, in every case t will be (2q^2
- 1) q^2.
Thus, the next triangular then will be q^2 (2q^2 + 1).
And this is a product of three primes
if and only if (2q^2 + 1) is a prime : i. e. : p and
p + 2 must be twin primes.
And this is only possible when q = 3.
Summarizing, we have proved that, if t, t' are two consecutive
triangular numbers with only three prime factors,
then t = 17*9 and t = 9*19, Q. E. D.
***
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